Answer
The initial acceleration of the 3rd sphere is $2.25\times10^{-10}m/s^2$
Work Step by Step
1) Take the mass of sphere 1 and sphere 2 to be $M=2.8kg$ and sphere 3 to be $m$
The gravitational force that sphere 1 and sphere 2 each has exerted on sphere 3 is $$F_{13}=F_{23}=G\frac{M\times m}{r^2}$$
Now we need to find the net force on sphere 3.
As we can see in the below image, since $F_{13}=F_{23}$, the horizontal forces $F_{13}\sin\theta$ and $F_{23}\sin\theta$ cancel out each other.
So the net force on sphere 3 is the sum of only the vertical component of $F_{13}$ and $F_{23}$ $$\sum F_3=F_{13}\cos30+F_{23}\cos30=2G\frac{Mm}{r^2}\cos30$$ $$\sum F_3=1.73G\frac{Mm}{r^2}$$
We know $G=6.67\times10^{-11}Nm^2/kg^2$, $M=2.8kg$ and distance $r=1.2m$
$$\sum F_3=(2.25\times10^{-10}m)N$$
2) From Newton's 2nd Law: $$\sum F_3=ma_3=(2.25\times10^{-10}m)N$$ $$a_3=2.25\times10^{-10}m/s^2$$
